Question

A block of pine wood (p=.43 g/cm³) is floating on a pond. The block is 10 cm x 40 cm x 5 cm thick.

a) How much of the block protrudes above the water?

b) If the block is made to carry a load by placing additional mass on top of it, how much mass must be added to just submerge the block?

Please show all work. Thank you for the help!!!

Answer 1

The relevant physics is as follows:
When a solid floats or is immersed in a fluid, it experiences two principal forces: (i) its weight, which is the product of its mass and the acceleration due to gravity (g); and (ii) a buoyancy force, which is equal to the weight of the fluid that is displaced by the solid's presence.

Say the solid has a mass m and a (total) volume v_tot. The weight of the object will be simply m*g. You say that the object floats, i.e. not all of its volume is submerged. Let's say that a volume v_sub is submerged. The volume of fluid displaced is therefore v_sub. The mass of fluid displaced is going to be v_sub * d_fluid, where d_fluid is the density of the fluid. Now, the weight of the fluid displaced is going to be v_sub * d_fluid * g, which will be the force of buoyancy.

Now, if the object is floating still (not sinking), the net force on it is zero, i.e. its weight and the buoyancy force are equal and opposite. Therefore,

m * g = v_sub * d_fluid * g, [1]

and the g will cancel, leaving

m = v_sub * d_fluid,

which can be rearranged for v_sub as

v_sub = m / d_fluid.

Now, the volume of the object that is above the fluid will be v_above = v_tot - v_sub, and from the previous equation, this will be

v_above = v_tot - (m / d_fluid).

The mass of the block will be its volume (v_tot) times its density (d_block). Therefore,

v_above = v_tot - (v_tot * d_block / d_fluid),

and pulling v_tot out as a factor, we get

v_above = v_tot * (1 - (d_block / d_fluid)).

If you want to express the fraction of volume that is protruding, then you want (v_above / v_tot), which from the previous result will be

(v_above / v_tot) = 1 - (d_block / d_fluid).

Now, you have d_block, and d_water is easy to remember at about 1000 kg/m^3 (i.e. 1 g/cm^3), so you can quickly give the fraction of the block that floats above the water as 1 - (d_block / d_water).

i.e 1- (0.43/1) = 0.57

For the second part, you need to add the additional weight to the left-hand side of equation [1], and solve in a similar way, noting that v_sub = v_tot (the block is completely submerged).

m* =mass needed to submerge the wood

m*g=v_sub* d_block* g

m*= v_tot * d_block

m*= 2000* 1 =2000g or 2kg